Coin Flip Probabilities: One, Two, and n Coins
A fair coin is flipped. What is the probability of heads?
Now flip two fair coins. What is the probability of getting at least one head? What is the probability of exactly one head?
Finally, generalize to $n$ fair coin flips. What is the distribution of the number of heads? Give the probability of exactly $k$ heads, at least one head, and describe the distribution for large $n$.
Hints
- For two coins, enumerate all four outcomes and count. For the general case, think about what distribution governs the count of successes in $n$ independent Bernoulli trials.
- The number of heads in $n$ fair flips follows $\text{Binomial}(n, 1/2)$. The probability of exactly $k$ heads is $\binom{n}{k} / 2^n$.
- For large $n$, the CLT tells you the sum of i.i.d. random variables is approximately normal. For $X \sim \text{Binomial}(n, 1/2)$: mean $= n/2$, variance $= n/4$, so $X \approx N(n/2, n/4)$.
Worked Solution
How to Think About It: This is a warm-up problem that tests whether you can set up the binomial distribution cleanly and know the large-$n$ normal approximation. For a single coin, everything is trivial. For $n$ coins, the key tool is the Binomial$(n, 1/2)$ distribution -- each flip is independent, each has the same probability of heads, and you count the successes.
Quick Estimate: For large $n$ (say $n = 100$), the expected number of heads is $50$ with standard deviation $\sqrt{100 \times 0.5 \times 0.5} = 5$. By CLT, the number of heads is approximately $N(50, 5^2)$, so about 68% of the time you get between 45 and 55 heads.
Formal Solution:
1 coin: $P(\text{heads}) = \frac{1}{2}$
2 coins:
All outcomes: $\{HH, HT, TH, TT\}$, each equally likely with probability
$P(\text{at least 1 head}) = 1 - P(\text{no heads}) = 1 - \frac{1}{4} = \frac{3}{4}$
$P(\text{exactly 1 head}) = P(HT) + P(TH) = \frac{2}{4} = \frac{1}{2}$
$n$ coins:
The number of heads $X \sim \text{Binomial}(n, 1/2)$. By the binomial theorem:
$P(X = k) = \binom{n}{k} \left(\frac{1}{2}\right)^n, \quad k = 0, 1, \ldots, n$
$P(X \geq 1) = 1 - P(X = 0) = 1 - \left(\frac{1}{2}\right)^n$
Moments: $E[X] = \frac{n}{2}, \qquad \text{Var}(X) = \frac{n}{4}$
Large $n$ (CLT approximation): $X \approx N\!\left(\frac{n}{2},\, \frac{n}{4}\right)$
This approximation is good when $n \geq 30$ or so. To compute $P(X \geq t)$ for a specific threshold $t$, standardize: $Z = \frac{X - n/2}{\sqrt{n}/2}, \quad P(X \geq t) \approx 1 - \Phi\!\left(\frac{t - n/2}{\sqrt{n}/2}\right)$
Answer: - 1 coin: $P(H) = 1/2$. - 2 coins: $P(\geq 1H) = 3/4$, $P(\text{exactly } 1H) = 1/2$. - $n$ coins: $X \sim \text{Binomial}(n, 1/2)$, $P(k \text{ heads}) = \binom{n}{k}/2^n$, $P(\geq 1H) = 1 - 1/2^n$, and for large $n$, $X \approx N(n/2, n/4)$.
Intuition
The binomial distribution is the workhorse of discrete probability, and the fair coin ($p = 1/2$) version is the cleanest case. The key facts to internalize: (1) probabilities are symmetric around $n/2$; (2) the spread grows as $\sqrt{n}$ -- the distribution gets wider in absolute terms but narrower *relative* to $n$; (3) for large $n$, the CLT kicks in and you can replace the exact binomial with a Gaussian.
In quant finance, this shows up in win-rate analysis (how many profitable trading days out of $n$?), in option pricing (the binomial tree is literally the Binomial distribution in disguise), and in hypothesis testing (is this strategy's win rate above 50%?). The key practical skill is knowing when the Gaussian approximation is good enough -- and it usually is for $n > 30$, which covers most real-world cases where you are looking at weekly or monthly trade counts.